calculate-0-pi-2-d-1-2sin-2-

Question Number 36201 by prof Abdo imad last updated on 30/May/18

c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + 2 {s i n}^{2} θ}

Commented by prof Abdo imad last updated on 01/Jun/18

l e t p u t I = \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + 2 {s i n}^{2} θ}

I = \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + 2 \frac{1 - c o s (2 θ)}{2}} = \int_{0}^{\frac{π}{2}} \frac{d θ}{2 - c o s (2 θ)}

=_{2 θ = t} \int_{0}^{π} \frac{1}{2 - c o s (t)} \frac{d t}{2}

=_{t a n (\frac{t}{2}) = x} \int_{0}^{+ \infty} \frac{1}{2 {2 - \frac{1 - x^{2}}{1 + x^{2}}}} \frac{2 d x}{1 + x^{2}}

= \int_{0}^{+ \infty} \frac{d x}{2 (1 + x^{2}) - 1 + x^{2}}

= \int_{0}^{+ \infty} \frac{d x}{1 + 3 x^{2}} c h a n g e m e n t x \sqrt{3} = u g i v e

I = \int_{0}^{\infty} \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{3}} = \frac{1}{\sqrt{3}} \frac{π}{2} \Rightarrow I = \frac{π}{2 \sqrt{3}} .